What is the solution for y^{''}-64y=16 ?

The solution for y^{''}-64y=16 is y=c_{1}e^{8t}+c_{2}e^{-8t}-1/4

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y^{''}-64y=16

y^{^{''}} - 64 y = 16

y = c_{1} e^{8 t} + c_{2} e^{- 8 t} - \frac{1}{4}

Solution steps

y^{''} (t) - 64 y = 16

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- 8 t} - \frac{1}{4}

y = c_{1} e^{8 t} + c_{2} e^{- 8 t} - \frac{1}{4}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = sin (e^{x})

y^{^{''}} + 3 y^{^{'}} - 10 y = x (e^{2 x} + 1)

\frac{d ^{2} y}{d x ^{2}} - y = cosh (x)

y^{^{''}} + 9 y = x e^{2 x} + sin (3 x)

y^{^{''}} - 4 y^{^{'}} - 12 y = 2 t^{3} - t + 3

What is the solution for y^{''}-64y=16 ?
The solution for y^{''}-64y=16 is y=c_{1}e^{8t}+c_{2}e^{-8t}-1/4